Since, point A (sec$$\theta$$, 2tan$$\theta$$) lies on the hyperbola 2x² $$-$$ y² = 2

Therefore, 2sec²$$\theta$$ $$-$$ 4tan²$$\theta$$ = 2

$$\Rightarrow$$ 2 + 2tan²$$\theta$$ $$-$$ 4tan²$$\theta$$ = 2

$$\Rightarrow$$ tan$$\theta$$ = 0 $$\Rightarrow$$ $$\theta$$ = 0

Similarly, for point B, we will get $$\phi$$ = 0.

but according to question $$\theta$$ + $$\phi$$ = $${\pi \over 2}$$ which is not possible.

Hence, it must be a 'BONUS'.

Tangent

$${{x\cos \theta } \over b} + {{y\sin \theta } \over {2a}} = 1$$



So, area $$(\Delta OAB) = {1 \over 2} \times {b \over {\cos \theta }} \times {{2a} \over {\sin \theta }}$$

$$ = {{2ab} \over {\sin 2\theta }} \ge 2ab$$

$$\Rightarrow$$ k = 2

Given C(3, $$-$$4), S(4, $$-$$4)



and A(5, $$-$$4)

Hence, a = 2 & ae = 1

$$\Rightarrow$$ e = $${1 \over 2}$$

$$\Rightarrow$$ b² = 3

So, $$E:{{{{(x - 3)}^2}} \over 4} + {{{{(y + 4)}^2}} \over 3} = 1$$

Intersecting with given tangent.

$${{{x^2} - 6x + 9} \over 4} + {{{m^2}{x^2}} \over 3} = 1$$

Now, D = 0 (as it is tngent)

So, 5m² = 3.

Let, $$P \equiv \left( {{3 \over 2}{t^2},3t} \right)$$ is on the curve.

Normal at point P

$$tx + y = 3t + {3 \over 2}{t^3}$$

Passes through $$\left( {3,{3 \over 2}} \right)$$

$$ \Rightarrow 3t + {3 \over 2} = 3t + {3 \over 2}{t^3}$$

$$ \Rightarrow {t^3} = 1 \Rightarrow t = 1$$

$$P \equiv \left( {{3 \over 2},3} \right) = (\alpha ,\beta )$$

$$2(\alpha + \beta ) = 2\left( {{3 \over 2} + 3} \right) = 9$$